We want to build an oil pipeline between two wells A and B located in opposite vertices of a rectangular shaped wetland of sides a and b (a < b).
The price per meter built on dry land is $c_1$ euros, while on the wetland it is $c_2 > c_1$ euros.
To optimize the price, the engineers in charge of the project consider, (besides building the oil pipeline just going through the wetland or only on dry land bordering it) the trajectories from the figure below, where the length built on dry land varies.
Find a trajectory that minimizes the cost, and the minimum costs depending on the parameters $a,b,c_1, c_2$
Don’t know where to start, any help or hints are very much appreciated
So it would benefit you if you established a point of origin. Let’s choose the bottom, right-hand corner.
Now notice if you have a trajectory entirely in the bottom diagonal (everything below a line from A to B), it would have a symmetric trajectory in the top diagonal! Moreover, these trajectories would be evaluated as having the same price. This is a way of simplifying your search domain. But you technically need to prove that no such trajectory that spans both the top and bottom regions could optimize the problem.
So now that you’ve split the domain, you can write an effective cost function. Given a path where you go from:
You obtain the cost function: $$f(x,y) = c_1(A-x) + c_2\sqrt{x^2+y^2}+c_1(B-y)$$
Notice that if we let $x,y = A,B$, you get the entirety of the pipe in the water, but if you let $x,y = 0,0$, the pipe stays strictly on dry land.
Now that we have specified the cost function, all you have to do is solve for all the critical points within the domain: $$0 \leq x \leq A \ \ \ 0 \leq y \leq B $$
Solving for the critical points, you see that solving the system of equations:
$$\frac{\partial f}{\partial x} = -c_1 + c_2 x (x^2+y^2)^{-1/2} = 0$$ $$\frac{\partial f}{\partial y} = -c_1 + c_2 y (x^2+y^2)^{-1/2} = 0$$
You obtain: $$\begin{cases} (c_2 - c_1) x^2 = c_1 y^2 \\ (c_2 - c_1) y^2 = c_1 x^2 \end{cases} \implies (c_2 - c_1) y^2 = \frac{c_1^2}{c_2 - c_1} y^2 \implies (c_2 - c_1)^2 y^2 = c_1^2 y^2$$
So we're left with the cases: $$y = 0 \implies x = 0 \implies (x,y) = (0,0)$$ $$y \not = 0 \implies c_2 - c_1 = c_1 \implies c_2 = 2 c_1 \implies x = \frac{c_1}{c_2 - c_1}y \implies (x,y) = (\frac{c_1}{c_2 - c_1}y,y) = f(y,y) \implies A = B$$
As well as the points on the domain where we lack differentiability: $$(A,B)$$ $$(x,B)$$ $$(A,y)$$
Thus, evaluating $f$ at each of the 5 cases, we see that: $$f(0,0) = c_1(A + B)$$ $$f(A,B) = c_2\sqrt{A^2 + B^2}$$ $$f(x,B) = c_1(A - x) + c_2\sqrt{x^2 + B^2}$$ $$f(A,y) = c_1(B - y) + c_2\sqrt{A^2 + y^2}$$
And the special case where $c_2 = 2 c_1$ $$f(y,y) = c_1(A + B) - 2 c_1 y + c_2\sqrt{A^2 + B^2} \geq c_2\sqrt{A^2 + B^2}$$
So we're left with optimizing over the following cases: $$\begin{cases} f(0,0) = c_1(A + B) \\ f(A,B) = c_2\sqrt{A^2 + B^2} \\ f(x,B) = c_1(A - x) + c_2\sqrt{x^2 + B^2} \\ f(A,y) = c_1(B - y) + c_2\sqrt{A^2 + y^2} \end{cases}$$
which is a one-dimensional optimization problem since $x,y$ nonzero imply the other zero. This is actually evident in the drawing you provided.